# Putnam 2012 A6

• Dec 14th 2012, 07:54 PM
skeptopotamus
Putnam 2012 A6
It's a nice problem; I have no idea how to solve it, though, and cannot find any solutions online.

Let $f$ be a continuous function on $\mathbb{R}^2$ such that the double integral of $f$, taken over any rectangle of area one, is zero. Does this necessarily imply that $f$ is identically zero?

As far as I recall, that is the precise wording of the question; there is not elaboration on what type of integral or what type of continuity it exhibits, but because it is on the Putnam I think the interpretation should be Riemann double integral, and absolute continuity.

Intuitions? Thoughts about a solution? IMO it seems to be true, but have no idea how to prove. Possibly an argument from Fourier, but I don't want to mess with it.
• Dec 14th 2012, 11:06 PM
chiro
Re: Putnam 2012 A6
Hey skeptopotamus.

What if the function is symmetrical with a period of exactly 1?

For example z = f(x,y) = sin(2*pi*x)*cos(2*pi*y).

Try integrating that over any simple rectangle of unit area and see if its zero.
• Dec 14th 2012, 11:30 PM
skeptopotamus
Re: Putnam 2012 A6
Hi, thanks for the suggestion. My feeling is that if there is a counterexample, it will be a similar sort of thing. Unfortunately, for the function you describe above, integrate over 2 < x < 3, 3 < y < 4. Also, I don't think that the rectangles have to be simple.

Wolfram|Alpha Widgets: "Double Integral Calculator" - Free Mathematics Widget
• Dec 14th 2012, 11:33 PM
chiro
Re: Putnam 2012 A6
Try making it z = cos(2*pi*x)*cos(2*pi*y) or z = sin(2*pi*x)*sin(2*pi*y)
• Dec 14th 2012, 11:42 PM
skeptopotamus
Re: Putnam 2012 A6
With those two I can take $I^2$ and rotate by epsilon to get a non-zero double integral.
• Dec 14th 2012, 11:44 PM
chiro
Re: Putnam 2012 A6
I should have asked this, can the unit rectangle be non-aligned with the primary x and y axes?
• Dec 14th 2012, 11:55 PM
skeptopotamus
Re: Putnam 2012 A6
Yes, or at least the question did not preclude it. Although I'm not sure that it would matter in the end (although it may be that the easiest proof in the positive, were it to be the case, utilizes this).